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Let $E$ be a commutative monoid with identity $e$, $S\subseteq E$, and let $S'$ be the submonoid of E generated by $S$.

For every invertible element $x$ of E, $x^{*}$ denotes the inverse of $x$.

Let $R(x,y)$ denote the following relation:

$$\text{there exist }a,b\in E\text{ and }p,q,s\in S'\text{ such that }x=(a,p), y=(b,q)\text{ and }aqs=bps;$$

this is an equivalence relation on the product monoid $E\ \times\ S'$, compatible with its law. Denote the quotient monoid $E\times S'/R$ by $E_{S}$.

For $a\in E$ and $p\in S'$, let $a/p$ denote the equivalence class of $(a,p)$ modulo R. Then, mapping

$$\varepsilon: E\longrightarrow E_{S},\ a\mapsto a/e,$$

is a homomorphism.

For every $a\in E$ and $p\in S'$, we have $a/p=(a/e)(e/p)$: i.e., $a/p=\varepsilon(a)\varepsilon(p)^{*}.$

Therefore, $\varepsilon(E) \cup\varepsilon(S)^{*}$ generates $E_{S}$, where $\varepsilon(S)^{*}$ denotes the set of invertible elements of $\varepsilon(S)$.

This construction is taken from Bourbaki's Algebra. Now Bourbaki goes on to say: enter image description here

Question: Can someone please explain what is going on in this screenshot? (I understand why $\varepsilon$ is injective.) In particular, I don't understand how one can "identify" $E_S$ with the submonoid of $\Phi$ generated by $E\cup S^*$? This whole business of "identification" is unclear to me: I don't understand what I am exactly able to do logically and formally whenever I am to told to "identify" two objects.

Any help will be appreaciated. Thank you.

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  • $\begingroup$ How exactly does this question differ from your previous one, which was also asking about "identifying" ? $\endgroup$ Dec 16, 2019 at 20:00
  • $\begingroup$ Well, it has more information which will hopefully result in a better answer.---I am myself aware of the fact; therefore, I am not sure what you hope to achieve by your remark. If this question weighs greatly on your conscience, you may of course vote to close it or downvote etc. etc.--- I don't appreciate your tone: your internet points do not give you the right to speak down to people. $\endgroup$
    – spring
    Dec 16, 2019 at 20:17
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    $\begingroup$ There was no tone: there was a question. Yours is a long question, it seems to cover the same ground, it includes an identical gif from a text. If you read that as accusatory, then you misread it. If you only wanted to add more information, then best practice is to edit your previous question to add it, not to post a nearly identical question with extra detail. As to "speak down to people", again, you are incorrect. You are relatively new to the site, and I am more familiar with it; I am trying to guide you to best practices. If the query so offended you, perhaps you were feeling guilty? $\endgroup$ Dec 16, 2019 at 20:23
  • $\begingroup$ Moreover, as you can see, my voting to close your question as duplicate results in an automatic closure, so before I did so I wanted to know if I missed something. Given that you say I did not, well... $\endgroup$ Dec 16, 2019 at 20:24

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